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A drawer holds 4 red hats and 4 blue hats. What is the

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A drawer holds 4 red hats and 4 blue hats. What is the [#permalink] New post 29 Aug 2004, 12:24
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A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer?

(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12
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 [#permalink] New post 29 Aug 2004, 18:56
Hmm, anybody else got 4/35?
4/8 * 3/7 * 2/6 * 4/5 * 2 = 4/35
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 [#permalink] New post 29 Aug 2004, 19:05
the OA is 1/4 but i dont know how to do this problem
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 [#permalink] New post 29 Aug 2004, 21:39
We can either end up with 3 red hats and 1 blue hat, or 3 blue hats and 1 red hats

Using hypergeometric distribution,
probability of getting 3 red hat and 1 blue hat = probability of getting 3 blue hat and 1 red hat
P = 2(4C3 * 4C1/(4+4)C(3+1)) = 2(8/35) = 16/35 ??
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 [#permalink] New post 30 Aug 2004, 06:05
I got 3/8.

P(3 RED + 1 BLUE) + P(3 BLUE + 1 RED): = 3/4 x 1/4 + 3/4 x 1/4
= 2(3/16) = 3/8

Does this make sense, or am I missing something ?
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 [#permalink] New post 30 Aug 2004, 11:16
ChallengeBound wrote:
I got 16/35's.

((4C3 * 4) * 2 ) / 8C4


I understand 4C3 * 4 * 2, but could you explain the 8C4? How would that change if there were only 3 red hats?
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 [#permalink] New post 30 Aug 2004, 11:21
jhs wrote:
ChallengeBound wrote:
I got 16/35's.

((4C3 * 4) * 2 ) / 8C4


I understand 4C3 * 4 * 2, but could you explain the 8C4? How would that change if there were only 3 red hats?

8C4 is total number of way to pick 4 hats out of 8. Also, I agree with 16/35 but I'm just wondering how my intuitive approach gave me 4/35... Anybody cares to explain? Sometimes, these subtle differences still elude me :?
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 [#permalink] New post 30 Aug 2004, 11:34
People how about 3/8 like I have explained earlier. Does it seem right ?

P(3 RED + 1 BLUE) + P(3 BLUE + 1 RED): = 3/4 x 1/4 + 3/4 x 1/4
= 2(3/16) = 3/8

Can somebody please ponder on this solution ? Does it seem right ?
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 [#permalink] New post 30 Aug 2004, 12:19
Paul wrote:
jhs wrote:
ChallengeBound wrote:
I got 16/35's.

((4C3 * 4) * 2 ) / 8C4


I understand 4C3 * 4 * 2, but could you explain the 8C4? How would that change if there were only 3 red hats?

8C4 is total number of way to pick 4 hats out of 8. Also, I agree with 16/35 but I'm just wondering how my intuitive approach gave me 4/35... Anybody cares to explain? Sometimes, these subtle differences still elude me :?


I think when doing 4C3 you must have divided by 4. thats what I did. First I got 16/35 and when the answer was not there I tried to redo and got 4/35.
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 [#permalink] New post 30 Aug 2004, 14:48
amit_drummer wrote:
People how about 3/8 like I have explained earlier. Does it seem right ?

P(3 RED + 1 BLUE) + P(3 BLUE + 1 RED): = 3/4 x 1/4 + 3/4 x 1/4
= 2(3/16) = 3/8

Can somebody please ponder on this solution ? Does it seem right ?


I can't follow this logic, but here's how I first interpreted the problem.
I used this dummy logic:

RRRB + RBRR + BRRR + RRBR + BBBR + BRBB + RBBB + BBRB
= 1/2 * 3/7 * 2/6 * 4/5 * 8
= 2/35 * 8
= 16/35

I still have trouble translating the problem in terms of combinations though...
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 [#permalink] New post 30 Aug 2004, 18:51
Possible outcomes = RRRR, RRRB, RRBB, RBBB, BBBB. Therefore in any way he picks the 4 hats, he get 5 possible outcomes. And in these 5, there are only 2 outcomes which have exactly 3 hats of same color. (the order does not matter)

Hence. P=2/5

Is this the right way?
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 [#permalink] New post 30 Aug 2004, 21:15
i think we all are coming at different answers, because the Question doesn't clearly mention, that the hats are drawn after replacement, or without replacement, i.e. if 1 hat is drawn, is it put back in the drawer to draw the second, or the 2nd one is drawn from the remaining 7 hats.

lets for a moment assume, that the hats are drawn after replacement.

so, the prob of drawing a red hat in any trial = 1/2.
so the prob of drawing exactly 3 red and 1 blue in 4 trials is:

4C3 * (1/2)^3 * (1/2) ^1 = 1/4. the same applies when it is 3blues and 1 red. so the combined probabililty is 2* 1/4 = 1/2.



going by the other assumption, that the hats are not replaced back...
the ways in which it can be drawn is RRRB, RRBR, RBRR, BRRR. and the prob. of this is:

4/8*3/7*2/6*4/5 + 4/8*3/7*4/6*2/5 + 4/8*4/7*3/6*2/5 + 4/8*4/7*3/6*3/5 =..... (some crap no. which i dont want to calculate) .
and the desired probability is double of this number, becasue it can 3blue hats and 1 red, which will also be same as this.
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 [#permalink] New post 30 Aug 2004, 21:23
It is clear from the question that it is without replacement. How can one have 4 hats in hand if he replaces the hat everytime to take the next one. IT says, Take 4 and check if there are 3 of same color. AM i right?
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 [#permalink] New post 30 Aug 2004, 21:31
Now, to point out some thing which most of us missed is :

soln 1, from Paul: 4/8 * 3/7 * 2/6 * 4/5 * 2 = 4/35 (this assumes that the hats are not replaced back in te drawer, but at the same time misses the order in which the hats are drawn.. it assumes the order to be RRRB and BBBR) :?

soln2, from ChallengeBound: ((4C3 * 4) * 2 ) / 8C4 = 16/35. (this baffles me completely, because, it probably assumes all the 8 hats are different, that is whether i pick hat1 of red color or hat2 of red color, logically it shud be same, but by taking total ways of selections of hat, it assumes all the 8 hats are different) :?

soln3, from amit_drummer: P(3 RED + 1 BLUE) + P(3 BLUE + 1 RED): = 3/4 x 1/4 + 3/4 x 1/4 (this one assumes that the hats are drawn after replacement, but again misses the order in which they are drawn) :?

soln4: from carsen: RRRR, RRRB, RRBB, RBBB, BBBB: so it shud be 2/5, this soln obviously doesn't consider the individual probabilities of picknig up the hats... :?


OA = 1/4 as pointed by damit: doesn't convince me, as by taking all the points into consideration i conclude answer to be 1/2... (explained in my earlier message)
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 [#permalink] New post 30 Aug 2004, 21:47
hi target

Why shud it consider individual probs of the hat. I repeat again, the person has to pick 4 out of the 8 and check if he has 3 of the same color (i am putting it in simple words).

The question asked is , what is the probability that 3 of the 4 are of same color. why do we need to find individual probs.

Say, I have a box with 8 balls. Of which 4 are blue, and 4 are green. I ask someone to put his hand inside and pick the 4 balls. Now, we can find out if the 3 balls are of same color only, and only after we see all the 4 of them. Right. Not that, the first one is blue or the order. The possible outcomes of one shot (picking up 4 balls) is as said before. RRRR, RRRB..

List out the possible combination of the 4 balls. And check which among the group have EXACTLY 3 of the same color. That's it.
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 [#permalink] New post 30 Aug 2004, 22:35
i think the hypergeometric progression approach i took is based on not replacing the hats. Is that correct ? Can anyone tell me if my approach is wrong. It does look correct, similar to one of those probability questions about having for instance, 6 balls, 3 red and 3 green and the probability of picking up 3 red and 1 green ball.
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 [#permalink] New post 31 Aug 2004, 04:50
Hi everyone,

I am puzzled by this question as I first of all wrote 2*4C3*4C1/8C4=16/35 but it suggests hats are different so wrong

In simply writing the 16 possible picks I obtain a 1/2 probability and I do not understant how to reach 1/4 ...
  [#permalink] 31 Aug 2004, 04:50
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